\(\int x \arccos (a x)^{5/2} \, dx\) [89]

Optimal result

Integrand size = 10, antiderivative size = 119 \[ \int x \arccos (a x)^{5/2} \, dx=\frac {15 \sqrt {\arccos (a x)}}{64 a^2}-\frac {15}{32} x^2 \sqrt {\arccos (a x)}-\frac {5 x \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{8 a}-\frac {\arccos (a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \arccos (a x)^{5/2}+\frac {15 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arccos (a x)}}{\sqrt {\pi }}\right )}{128 a^2} \] Output:

15/64*arccos(a*x)^(1/2)/a^2-15/32*x^2*arccos(a*x)^(1/2)-5/8*x*(-a^2*x^2+1) 
^(1/2)*arccos(a*x)^(3/2)/a-1/4*arccos(a*x)^(5/2)/a^2+1/2*x^2*arccos(a*x)^( 
5/2)+15/128*Pi^(1/2)*FresnelC(2*arccos(a*x)^(1/2)/Pi^(1/2))/a^2
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.61 \[ \int x \arccos (a x)^{5/2} \, dx=\frac {15 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arccos (a x)}}{\sqrt {\pi }}\right )-2 \sqrt {\arccos (a x)} \left (\left (15-16 \arccos (a x)^2\right ) \cos (2 \arccos (a x))+20 \arccos (a x) \sin (2 \arccos (a x))\right )}{128 a^2} \] Input:

Integrate[x*ArcCos[a*x]^(5/2),x]
 

Output:

(15*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcCos[a*x]])/Sqrt[Pi]] - 2*Sqrt[ArcCos[a*x] 
]*((15 - 16*ArcCos[a*x]^2)*Cos[2*ArcCos[a*x]] + 20*ArcCos[a*x]*Sin[2*ArcCo 
s[a*x]]))/(128*a^2)
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {5141, 5211, 5141, 5153, 5225, 3042, 3793, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x \arccos (a x)^{5/2} \, dx\)

\(\Big \downarrow \) 5141

\(\displaystyle \frac {5}{4} a \int \frac {x^2 \arccos (a x)^{3/2}}{\sqrt {1-a^2 x^2}}dx+\frac {1}{2} x^2 \arccos (a x)^{5/2}\)

\(\Big \downarrow \) 5211

\(\displaystyle \frac {5}{4} a \left (\frac {\int \frac {\arccos (a x)^{3/2}}{\sqrt {1-a^2 x^2}}dx}{2 a^2}-\frac {3 \int x \sqrt {\arccos (a x)}dx}{4 a}-\frac {x \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{2 a^2}\right )+\frac {1}{2} x^2 \arccos (a x)^{5/2}\)

\(\Big \downarrow \) 5141

\(\displaystyle \frac {5}{4} a \left (-\frac {3 \left (\frac {1}{4} a \int \frac {x^2}{\sqrt {1-a^2 x^2} \sqrt {\arccos (a x)}}dx+\frac {1}{2} x^2 \sqrt {\arccos (a x)}\right )}{4 a}+\frac {\int \frac {\arccos (a x)^{3/2}}{\sqrt {1-a^2 x^2}}dx}{2 a^2}-\frac {x \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{2 a^2}\right )+\frac {1}{2} x^2 \arccos (a x)^{5/2}\)

\(\Big \downarrow \) 5153

\(\displaystyle \frac {5}{4} a \left (-\frac {3 \left (\frac {1}{4} a \int \frac {x^2}{\sqrt {1-a^2 x^2} \sqrt {\arccos (a x)}}dx+\frac {1}{2} x^2 \sqrt {\arccos (a x)}\right )}{4 a}-\frac {\arccos (a x)^{5/2}}{5 a^3}-\frac {x \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{2 a^2}\right )+\frac {1}{2} x^2 \arccos (a x)^{5/2}\)

\(\Big \downarrow \) 5225

\(\displaystyle \frac {5}{4} a \left (-\frac {3 \left (\frac {1}{2} x^2 \sqrt {\arccos (a x)}-\frac {\int \frac {a^2 x^2}{\sqrt {\arccos (a x)}}d\arccos (a x)}{4 a^2}\right )}{4 a}-\frac {\arccos (a x)^{5/2}}{5 a^3}-\frac {x \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{2 a^2}\right )+\frac {1}{2} x^2 \arccos (a x)^{5/2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5}{4} a \left (-\frac {3 \left (\frac {1}{2} x^2 \sqrt {\arccos (a x)}-\frac {\int \frac {\sin \left (\arccos (a x)+\frac {\pi }{2}\right )^2}{\sqrt {\arccos (a x)}}d\arccos (a x)}{4 a^2}\right )}{4 a}-\frac {\arccos (a x)^{5/2}}{5 a^3}-\frac {x \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{2 a^2}\right )+\frac {1}{2} x^2 \arccos (a x)^{5/2}\)

\(\Big \downarrow \) 3793

\(\displaystyle \frac {5}{4} a \left (-\frac {3 \left (\frac {1}{2} x^2 \sqrt {\arccos (a x)}-\frac {\int \left (\frac {\cos (2 \arccos (a x))}{2 \sqrt {\arccos (a x)}}+\frac {1}{2 \sqrt {\arccos (a x)}}\right )d\arccos (a x)}{4 a^2}\right )}{4 a}-\frac {\arccos (a x)^{5/2}}{5 a^3}-\frac {x \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{2 a^2}\right )+\frac {1}{2} x^2 \arccos (a x)^{5/2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {5}{4} a \left (-\frac {\arccos (a x)^{5/2}}{5 a^3}-\frac {3 \left (\frac {1}{2} x^2 \sqrt {\arccos (a x)}-\frac {\frac {1}{2} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arccos (a x)}}{\sqrt {\pi }}\right )+\sqrt {\arccos (a x)}}{4 a^2}\right )}{4 a}-\frac {x \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{2 a^2}\right )+\frac {1}{2} x^2 \arccos (a x)^{5/2}\)

Input:

Int[x*ArcCos[a*x]^(5/2),x]
 

Output:

(x^2*ArcCos[a*x]^(5/2))/2 + (5*a*(-1/2*(x*Sqrt[1 - a^2*x^2]*ArcCos[a*x]^(3 
/2))/a^2 - ArcCos[a*x]^(5/2)/(5*a^3) - (3*((x^2*Sqrt[ArcCos[a*x]])/2 - (Sq 
rt[ArcCos[a*x]] + (Sqrt[Pi]*FresnelC[(2*Sqrt[ArcCos[a*x]])/Sqrt[Pi]])/2)/( 
4*a^2)))/(4*a)))/4
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f 
, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
 

rule 5141

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[x 
^(m + 1)*((a + b*ArcCos[c*x])^n/(m + 1)), x] + Simp[b*c*(n/(m + 1))   Int[x 
^(m + 1)*((a + b*ArcCos[c*x])^(n - 1)/Sqrt[1 - c^2*x^2]), x], x] /; FreeQ[{ 
a, b, c}, x] && IGtQ[m, 0] && GtQ[n, 0]
 

rule 5153

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_S 
ymbol] :> Simp[(-(b*c*(n + 1))^(-1))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2] 
]*(a + b*ArcCos[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^ 
2*d + e, 0] && NeQ[n, -1]
 

rule 5211

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_. 
)*(x_)^2)^(p_), x_Symbol] :> Simp[f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + 
 b*ArcCos[c*x])^n/(e*(m + 2*p + 1))), x] + (Simp[f^2*((m - 1)/(c^2*(m + 2*p 
 + 1)))   Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcCos[c*x])^n, x], x] - S 
imp[b*f*(n/(c*(m + 2*p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p]   Int[(f* 
x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcCos[c*x])^(n - 1), x], x]) /; 
FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && IGtQ[m 
, 1] && NeQ[m + 2*p + 1, 0]
 

rule 5225

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 
2)^(p_.), x_Symbol] :> Simp[(-(b*c^(m + 1))^(-1))*Simp[(d + e*x^2)^p/(1 - c 
^2*x^2)^p]   Subst[Int[x^n*Cos[-a/b + x/b]^m*Sin[-a/b + x/b]^(2*p + 1), x], 
 x, a + b*ArcCos[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e 
, 0] && IGtQ[2*p + 2, 0] && IGtQ[m, 0]
 

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.66

method	result	size
default	\(\frac {32 \arccos \left (a x \right )^{\frac {5}{2}} \cos \left (2 \arccos \left (a x \right )\right ) \sqrt {\pi }-40 \arccos \left (a x \right )^{\frac {3}{2}} \sin \left (2 \arccos \left (a x \right )\right ) \sqrt {\pi }-30 \cos \left (2 \arccos \left (a x \right )\right ) \sqrt {\pi }\, \sqrt {\arccos \left (a x \right )}+15 \pi \,\operatorname {FresnelC}\left (\frac {2 \sqrt {\arccos \left (a x \right )}}{\sqrt {\pi }}\right )}{128 a^{2} \sqrt {\pi }}\)	\(79\)

Input:

int(x*arccos(a*x)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/128/a^2/Pi^(1/2)*(32*arccos(a*x)^(5/2)*cos(2*arccos(a*x))*Pi^(1/2)-40*ar 
ccos(a*x)^(3/2)*sin(2*arccos(a*x))*Pi^(1/2)-30*cos(2*arccos(a*x))*Pi^(1/2) 
*arccos(a*x)^(1/2)+15*Pi*FresnelC(2*arccos(a*x)^(1/2)/Pi^(1/2)))
 

Fricas [F(-2)]

Exception generated. \[ \int x \arccos (a x)^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x*arccos(a*x)^(5/2),x, algorithm="fricas")
 

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (constant residues)
 

Sympy [F]

\[ \int x \arccos (a x)^{5/2} \, dx=\int x \operatorname {acos}^{\frac {5}{2}}{\left (a x \right )}\, dx \] Input:

integrate(x*acos(a*x)**(5/2),x)
 

Output:

Integral(x*acos(a*x)**(5/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int x \arccos (a x)^{5/2} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(x*arccos(a*x)^(5/2),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.18 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.20 \[ \int x \arccos (a x)^{5/2} \, dx=\frac {\arccos \left (a x\right )^{\frac {5}{2}} e^{\left (2 i \, \arccos \left (a x\right )\right )}}{8 \, a^{2}} + \frac {\arccos \left (a x\right )^{\frac {5}{2}} e^{\left (-2 i \, \arccos \left (a x\right )\right )}}{8 \, a^{2}} + \frac {5 i \, \arccos \left (a x\right )^{\frac {3}{2}} e^{\left (2 i \, \arccos \left (a x\right )\right )}}{32 \, a^{2}} - \frac {5 i \, \arccos \left (a x\right )^{\frac {3}{2}} e^{\left (-2 i \, \arccos \left (a x\right )\right )}}{32 \, a^{2}} - \frac {\left (15 i + 15\right ) \, \sqrt {\pi } \operatorname {erf}\left (\left (i - 1\right ) \, \sqrt {\arccos \left (a x\right )}\right )}{512 \, a^{2}} + \frac {\left (15 i - 15\right ) \, \sqrt {\pi } \operatorname {erf}\left (-\left (i + 1\right ) \, \sqrt {\arccos \left (a x\right )}\right )}{512 \, a^{2}} - \frac {15 \, \sqrt {\arccos \left (a x\right )} e^{\left (2 i \, \arccos \left (a x\right )\right )}}{128 \, a^{2}} - \frac {15 \, \sqrt {\arccos \left (a x\right )} e^{\left (-2 i \, \arccos \left (a x\right )\right )}}{128 \, a^{2}} \] Input:

integrate(x*arccos(a*x)^(5/2),x, algorithm="giac")
 

Output:

1/8*arccos(a*x)^(5/2)*e^(2*I*arccos(a*x))/a^2 + 1/8*arccos(a*x)^(5/2)*e^(- 
2*I*arccos(a*x))/a^2 + 5/32*I*arccos(a*x)^(3/2)*e^(2*I*arccos(a*x))/a^2 - 
5/32*I*arccos(a*x)^(3/2)*e^(-2*I*arccos(a*x))/a^2 - (15/512*I + 15/512)*sq 
rt(pi)*erf((I - 1)*sqrt(arccos(a*x)))/a^2 + (15/512*I - 15/512)*sqrt(pi)*e 
rf(-(I + 1)*sqrt(arccos(a*x)))/a^2 - 15/128*sqrt(arccos(a*x))*e^(2*I*arcco 
s(a*x))/a^2 - 15/128*sqrt(arccos(a*x))*e^(-2*I*arccos(a*x))/a^2
 

Mupad [F(-1)]

Timed out. \[ \int x \arccos (a x)^{5/2} \, dx=\int x\,{\mathrm {acos}\left (a\,x\right )}^{5/2} \,d x \] Input:

int(x*acos(a*x)^(5/2),x)
 

Output:

int(x*acos(a*x)^(5/2), x)
 

Reduce [F]

\[ \int x \arccos (a x)^{5/2} \, dx=\int \sqrt {\mathit {acos} \left (a x \right )}\, \mathit {acos} \left (a x \right )^{2} x d x \] Input:

int(x*acos(a*x)^(5/2),x)
 

Output:

int(sqrt(acos(a*x))*acos(a*x)**2*x,x)